Problem: In triangle $ABC$, $AB = 9$, $BC = 12$, $AC = 15$, and $CD$ is the angle bisector.  Find the length of $CD$.
Solution: $\triangle ABC$ is a right triangle, since $9^2 + 12^2 = 15^2$. Thus, $\angle ABC = 90^\circ$.

[asy]
unitsize(0.3 cm);

pair A, B, C, D;

A = (0,9);
B = (0,0);
C = (12,0);
D = (0,4);

draw(A--B--C--cycle);
draw(C--D);

label("$A$", A, NW);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, W);
[/asy]

By the angle bisector theorem, $BD/AD = BC/AC$, so \[BD = \frac{BC}{BC + AC} \cdot AB = \frac{4}{9} \cdot 9 = 4.\] Then by use of the Pythagorean Theorem on right triangle $BCD$, $CD = \sqrt{BC^2 + BD^2} = \sqrt{12^2 + 4^2} = \sqrt{160} = \boxed{4 \sqrt{10}}$.